想要詳細步驟
Simplify
(n+3)x(n+2)!
a^2-1/(a+1)!
m!/(m^2-3m+2)
1/(m+1)!+1/(m-1)!
(a+1)!(b-1)!/(a-1)!(b+1)!
[數學]中五數學-概率問題
2011-11-05 11:56 pm
回答 (1)
2011-11-06 2:31 am
✔ 最佳答案
(n+3)x(n+2)!= (n+3)! a^2-1/(a+1)! = (a+1)(a-1)/(a+1)! = (a-1)/a! m!/(m^2-3m+2) = m!/(m-1)(m-2) = m(m-3)! 1/(m+1)!+1/(m-1)! = 1/(m+1)! + m(m+1)/(m+1)! = (m^2 + m +1)/(m+1)! (a+1)!(b-1)!/(a-1)!(b+1)! = a(a+1)/b(b+1)
收錄日期: 2021-04-13 18:20:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111105000051KK00596